An urn contains 2 red, 3 green and 2 blue balls. If 2 balls are drawn at random, find the probability that no ball is blue.
A. $$\frac{5}{7}$$
B. $$\frac{10}{21}$$
C. $$\frac{2}{7}$$
D. $$\frac{11}{21}$$
E. None of these
Answer: Option B
Solution (By Examveda Team)
Total number of balls = (2 + 3 + 2) = 7Let, E be the event of drawing 2 non-blue balls.
Then, n (E) = $${}^5\mathop C\nolimits_4 = \frac{{5 \times 4}}{{2 \times 1}}$$ = 10
And, n (S) = $${}^7\mathop C\nolimits_2 = \frac{{7 \times 6}}{{2 \times 1}}$$ = 21
$$\therefore P(E) = \frac{{n(E)}}{{n(S)}} = \frac{{10}}{{21}}$$
This Question Belongs to Arithmetic Ability >> Probability
5C4 = 5!/4!= 5
5C2 = 5!/2!*33!= 10
9/19 is the correct ans.